HBSE Class 12th Chemistry Solved Question Paper 2019
HBSE Class 12 Chemistry Previous Year Question Paper with Answer. HBSE Board Solved Question Paper Class 12 Chemistry 2019. HBSE 12th Question Paper Download 2019. HBSE Class 12 Chemistry Paper Solution 2019. Haryana Board Class 12th Chemistry Question Paper 2019 Pdf Download with Answer.
Q1.(i) What is tincture of iodine ?
(A) 0.2 to 0.4 ppm aqueous solution of chlorine
(B) 2-3% aqueous solution of CH3COOH
(C) 2-3% solution of iodine in alcoholwater
(D) None of the above
Ans. (C) 2-3% solution of iodine in alcoholwater
(ii) What is hybridization of Ni in [NiCl4]²- ?
Ans. (D) sp³
(iii) Which of the following organic compounds are formed by Wurtz reaction ?
Ans. (B) Hydrocarbons
(iv) Glycogen is an example of :
Ans. (A) Polysaccharide
(v) According to Hardy-Schulze rule, which of the following has highest flocculating power ?
(D) None of the above
Ans. (A) Al³+
(vi) What type of stoichiometric defect is shown by ZnS ?
Ans. Frenkel Defect
(vii) Give structure of monomer unit of Nylon-6.
(viii) What do you mean by Peptide linkage ?
Ans. It is linkage (or bond) by which α-amino acids are connected to each other in a protein.
(ix) Define Azeotrope mixture.
Ans. Azeotropes are binary mixtures having same composition in liquid and vapour phase and boil at constant temperature.
(x) Write the Nernst equation for following cell :
Sn(s) | Sn²+ || H+ | H2 (g)(1bar) | Pt(s)
Ans. E = E⁰ – 0.059/2 log [Sn²+]/[H+]²
(xi) Identify the reaction order from the following rate constant :
k = 3.5 × 10-⁵ L mol-¹ s-¹
Ans. Second Order
(xii) Write the reaction of Cl2 with water.
Ans. 2Cl2 + 2H2O → 4HCl + O2
Cl2 + H2O → 2HCl + [O]
(xiii) Write IUPAC name of (p) O2N–C6H4–OCH3
(xiv) Draw structure of cyclopropanone oxime.
Q2. If the density of some lake water is 1.25 g ml-¹ and contains 92 g of Na+ ions per Kg of water, calculate the molality of Na+ ions in the lake.
Ans. Mass of Na+ ions = 92 g
Moles of Na+ ions = 92/23 = 4
Mass of water = 1 Kg
Molality = 4/1 = 4 m
Q3. If a current of 0.5 ampere flows through a metallic wire for 2 hours, then how many electrons would flow through the wire ?
Ans. Coulombs of electrons = current × time = 0.5A × 2 × 60 × 60s = 3600 C
Now, 96500 C is equivalent to = 6.022 × 10²³ electrons
3600 C is equivalent to = (6.022×10²³ ÷ 96500) × 3600 = 2.246 × 10²² electrons
Q4. Differentiate between cationic and anionic detergents giving suitable examples.
Ans. Cationic detergents– These are quaternary ammonium salt of amines with acetates, chlorides or bromides. e.g. cetyltrimethyl ammonium bromide.
Anionic detergents– Anionic detergents are sod. salt of sulphonated long chain hydrocarbons or alcohols. e.g. Sod. laurylsulphate.
Q5. Explain the term copolymerization and give two examples.
Ans. Process in which two or more monomer combine to form a polymer. e.g.
(i) 1, 3-Butadiene and styrene
(ii) 1, 3-Butadiene and acrylonitrile
Q6. What do you mean by denaturation of a protein ? How does it affect properties of protein ?
Ans. When a protein in its native form is subjected to physical or chemical change, the hydrogen bonds are disturbed. This is called Denaturation of a protein.
Denaturation changes physical and biological properties but chemical properties remains the same.
Q7. It is safe to inject solutions isotonic with blood plasma intravenously. Explain.
Ans. During intravenous injections, the conc. of the solution to be injected should be comparable to blood plasma. If the solution is more conc. then water will flow out of cell & they would shrink. If the solution is less conc. then water will flow into cell & they would burst.
Q8. While giving labelled diagram of dry cell write reactions taking place at cathode and anode.
Anode : Zn(s) → Zn²+ + 2e-
Cathode : MnO2 + NH4+ + e- → MnO(OH) + NH3
Q9. Why does nitrogen show catenation properties less than phosphorus ?
Ans. Single N-N bond is weaker than single P-P bond because of high inter electronic repulsion of the non bonding electrons, owing to the small bond length. As a result the catenation tendency is weaker in Nitrogen.
Q10. Although amino group is o- and p- directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline. Give reason.
Ans. This is because in strongly acidic medium (conc. HNO3 + H2SO4), the aniline is pronated to form anilinium ion which is meta directing. Thus in addition to 0- & p- derivatives, significant amount of m-derivative is also formed.
Q11. Write a short note on Hofmann’s bromamide reaction.
Ans. This is a method for preparation of 1° amines in which an amide is treated with bromine in an aq. or alc. solution of NaOH.
R–C=O–NH2 + Br2 + 4NaOH → R–NH2 + Na2CO3 + 2NaBr + 2H2O
Q12.(a) What is meant by Hydroboration-oxidation reaction ? Illustrate it with an example.
Ans. Diborane (BH3)2 reacts with alkene to give trialkyl boranes as addition product. This is oxidized to alcohol by H2O2 in the presence of aq. NaOH. e.g.
CH3–CH=CH2 + (H–BH2)2 → CH3–CH2–CH2–BH2 + 2CH3CH=CH2 → (CH3–CH2–CH2)3B + 3H2O2/OH- + H2O → 3CH3–CH2–CH2–OH + B(OH)3
(b) Predict the major product of acid catalysed dehydration of 1-methylcyclohexanol.
Q13. Explain [Co(NH3)6]³+ is an inner orbital complex whereas [Ni(NH3)6]²+ is an outer orbital complex.
Ans. In the presence of strong ligand NH3, the 3d electrons pair up leaving two d orbitals to be involved in d²sp³ hybridization forming inner orbital complex.
In [Ni(NH3)6]²+ , Ni is in +2 0.s and has 3d⁸ configuration and hybridization involved is sp³d² forming outer orbital complex.
Q14. Define the following with suitable examples :
Ans. In metal excess defect some anion vacancies are created by passing alkali metal vapour over alkali metal crystal. Electrons are entrapped in these vacancies and responsible for importing colour to crystal.
Ans. If the alignment of magnetic moments of domain is in a compensatory way so as to give zero net mag. moment e.g. MnO.
Q15. Explain the following in context of solid catalysts :
Ans. It is the ability of catalyst to increase the rate of chemical reaction. e.g. H2 and O2 to form water in the presence of platinum.
2H2(g) + O2(g) + Pt → 2H2O(l)
Ans. It means the ability of the catalyst to direct reaction to give particular product e.g.
Co(g) + 3H2(g) + Ni → CH4(g) + H2O(g)
Q16. (a) The rate constants of a reaction at 500 K and 700 K are 0.02 S-¹ and 0.07 S-¹ respectively. Calculate the value of Ea.
Ans. log k2/k1 = Ea/2.303R [(T2-T1)/T2T1]
log 0.07/0.02 = (Ea/2.303×8.314) [(700-500)/700×500]
0.544 = Ea × 5.714 × 10⁴/19.15
Ea = 0.544×10-⁴×19.15/5.714 = 18230.8 J
(b) Under what condition a bimolecular reaction behaves kinetically first order reaction ?
Ans. A bimolecular Rx behaves kinetically 1st order Rx when one of the reactant is present in excess.
Q17. Why is the extraction of copper from pyrites more difficult than that from its oxide ore through reduction ?
Ans. The graph of ∆rG⁶ vs T in Ellingham diagram for the formation of oxides shows that the Cu-CuO line is almost at the top. Therefore it is very easy to reduce oxide ores of Cu directly to metal by heating with coke. This is because ∆rG⁶ vs T line for CO has –ve slope at higher temp. of therefore can easily reduce Cu2O to Cu.
However the Gibb’s energies for formation of most sulphides are greater than that for CS2. In fact CS2 is an endothermic compound. Moreover there is no CS analogous to CO. Therefore extraction of Cu from Cu2S is difficult.
Q18. Although chlorine is an electron withdrawing group, yet it is ortho-para directing in electrophilic aromatic substitution reactions, why ?
Ans. Chlorine withdraws electrons through inductive effect and releases electrons through resonance. Through inductive effect chlorine destabilizes the intermediate carbocation formed during E⊕ substitution.
Through resonance halogen tends to stabilize the carbocation and the effect is more pronounced at o- & p- position. The inductive effect is stronger than resonance effect & causes net deactivation. Reactivity is thus controlled by inductive effect and orientation is controlled by resonance effect.
Q19(a) How will you distinguish between Pentan-2-one and Pentan-3-one with the help of iodoform test ?
Ans. Pentan-2-one forms yellow ppt of iodoform with alkaline solution of iodine. Pentan-3one does not give this test
CH3COCH2CH2CH3 + 3I2 + 4NaOH → CH3CH2CH2COONa + CHI3 + 3H2O + 3NaI
(b) How will you bring about following conversions ?
(i) Benzoic acid to m-Nitrobenzyl alcohol
(ii)Benzaldehyde to Benzophenone
(iii) Benzoic acid to Benzamide
Describe the following :
(a) Aldol condensation
Ans. Aldehydes and ketones having at least one α-H undergo a reaction in the presence of dilute alkali as catalyst to form β-hydroxy aldehydes or βhydroxy ketones. The β-hydroxy aldehydes and ketones readily loose water to give α-β unsaturated carbonyl compounds. This Rx is called aldol condensation.
Ans. Carboxylic acids lose CO2 to form hydrocarbons when their sod. salt are heated with soda lime (NaOH + CaO) in the ratio of 3 : 1. The reaction is known as decarboxylation.
RCOONa + NaOH+Cao + Heat → R–H + Na2CO3
Q20.(a) Arrange the following in the order of property indicated for each set :
(i) F2, Cl2, Br2, I2 – Increasing bond dissociation enthalpy.
Ans. I2 < F2 < Br2 < Cl2
(ii) HF, HCl, HBr, HI – Decreasing acid strength.
Ans. HI > HBr > HCl > HF
(iii) NH3, PH3, AsH3, SbH3, BiH3 – Decreasing base strength.
Ans. NH3 > PH3 > AsH3 > SbH3 > BiH3
(b) Write the conditions to maximize the yield of H2SO4 by contact process.
Ans. Low temperature : But the temp. should not be very low otherwise rate of Rx will become slow. The optimum temp. 720 K should be maintained.
High pressure of 2 bar
Presence of Catalyst, V2O5
(a) Draw the structure of following :
(b) Describe the Haber manufacture of ammonia.
Ans. On large scale, ammonia is manufactured by Haber’s process.
N2(g) + 3H2(g) ⇌ 2NH3(g) ∆f4 = -46.1 KJ/Mol-¹
The optimum conditions for the production of NH3 are a pressure of 200 × 105 Pa, a temp. of ~700 K and the use of catalyst such as Ironoxide with small amount of K2O and Al2O3 to increase the rate of attainment at equilibrium.
Q21.(a) Why are Mn²+ compounds more stable than Fe²+ towards oxidation to their +3 state.
Ans. Electronic configuration of Mn²+ is [Ar]3d⁵ which is half filled and it is stable. It means third e- cannot be removed easily.
In case of Fe²+ , the electronic configuration is [Ar]3d⁶. Therefore one e- to acquire 3d⁵ stable electronic configuration.
(b) What are interstitial compounds ? Why are such compounds well-known for transition metals ?
Ans. interstitial compounds are those which are formed when small atoms like H, C or N are trapped inside crystal lattice of metals. The transition metals can easily accommodate these small atoms because of space between metal atoms in crystal lattice.
(c) Write electronic configuration of Pm³+ .
(a) Write ionic equations for the reaction of acidified potassium permanganate with following :
(i) Oxalic acid
(iii) Sulphite ion
(b) Describe the reactivity of actinoids.
Ans. The actinoids are highly reactive metals, especially when finely divided. The action of boiling water on them e.g. gives a mixture of oxide and hybride and combination with most non metals takes place at moderate temp. HCl attacks all metals but most are slightly affected by HNO3. Alkalies have no action.
Q1.(i) Which of the following Noble gas is used for advertisement display ?
Ans. (B) Ne
(ii) Which of the following is most acidic ?
Ans. Option (A)
(iii) Molal elevation constant is also called as :
(A) Cryoscopic constant
(B) Gas constant
(C) Ebullioscopic constant
(D) Freezing point depression constant
Ans. (C) Ebullioscopic constant
(iv) Example of semi-synthetic polymer is :
(A) Cellulose nitrate
Ans. (A) Cellulose nitrate
(v) CH3–CH2–OH + H2SO4 (413K)→X ; what is X ?
Ans. (B) C2H5–O–C2H5
(vi) Name two metals which are extracted electrolytically.
Ans. Cu, Ag, Ni, Zn, Al
(vii) What are complex reactions ?
Ans. When a sequence of elementary reactions give us the product, the reactions are called complex reactions.
(viii) Why is use of aspartame limited to cold foods and drinks ?
Ans. This is because aspartame is unstable at cooking temperature and decomposes.
(ix) What are reducing sugars ?
Ans. The sugars which can reduce Tollen’s reagent or Fehling solution are called reducing sugars.
(x) Name the disease caused by deficiency of Vitamin K.
Ans. Controls blood clotting time
(xi) What do you mean by Doping ?
Ans. Delibrately addition of impurities to a semiconductor to increase its conductivity is called doping.
(xii) Define coagulation.
Ans. The process of settling of colloidal particles is called coagulation.
(xiii) Write IUPAC name of K4[Mn(CN)6].
Ans. Potassium hexacyanomanganate (II)
(xiv) Give an example of β-elimination reaction of alkyl halides.
Q2. Write a short note on carbylamine reaction.
Ans. Aliphatic and aromatic 1° amines on heating with CHCl3 and ethanolic KOH form isocynide or carbylamine, is called carbyl amine Rx. e.g.
CH3CH2NH2 + CHCl3 + 3KOH + Heat → CH3CH2NC + 3KCl + 3H2O
Q3. How will you convert aniline into the following ?
Q4. Nitrogen exist as diatomic molecule and phosphorous as P4, why ?
Ans. N because of having small size & high e-vity is able to form Pπ-Pπ multiple bond & thus exist as N2.
On the other hand size of P is large it cannot form Pπ-Pπ multiple bond and it satisfy its valency by forming single bond & thus exist as P4.
Q5 Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Ans. (P°-P)/P° = x2 = WB/MB × MA/WA
(17.535-P)/17.535 = 25/180 × 18/450
17.535-P = 0.0975
P = 17.438 mm Hg
Q6. State Faraday’s first law of electrolysis.
Ans. The amount of a chemical compound deposited at any electrode during electrolysis is directly proportional to the quantity of electricity based.
W ∝ Q = W = ZQ
Q7. ∧°m for NaCl, HCl and NaAc are 126.4, 425.9 and 91.05 cm² mol-¹ respectively. Calculate ∧° for HAc.
Ans. ∧°m (HAc) = λ°H+ + λ°AC- = λ°H+ + λ°Cl- + λ°AC– + λ°Na+ – λ°Cl- – λ°Na+
= ∧°m(HCl) + ∧°m(NaAC) – ∧°m(NaCl)
= (425.9 + 91.0 – 126)S cm² mol-¹
= 390.5 S cm² mol-¹
Q8. With the help of Henry’s law, explain the medical condition known as bends.
Ans. Scuba divers cope with high conc. of dissolved gases while breathing air at high pressure under water. Increased pressure increases solubility of gases in blood. When the divers come towards surface, the pressure gradually decreases. This releases dissolved gases & lead to formation of bubbles of N2 in blood. This blocks capillaries and creates a medical condition known as bends.
Q9. What are Tranquilizers ? What is their use ?
Ans. Transquilizers are neurologically active drugs. These are used for treatment of stress and mild or even severe mental diseases.
Q10. Distinguish between the terms homopolymer and copolymer and give one example of each.
Ans. Homopolymer– It is a polymer which is formed from one type of monomer. e.g. polythene.
Copolymer– It is a polymer formed from two or more different monomers. e.g. Nylon-6,6.
Q11. Write four differences between DNA and RNA.
Ans. DNA– It has double stranded α-helix structure. Sugar molecule is 2-Deoxyribose. It has unique property. Nitrogeneous base uracil not present.
RNA– It has single stranded α-helix structure. Sugar molecule is ribose. RNA does not replicate. Nitrogeneous base thymine is not present.
Q12.(a) Copper crystallizes in fcc lattice with edge length 3.61 × 10-⁸ cm. Show that the calculated density is in agreement with its measured value of 8.92 g/cm³.
Ans. Density (ρ) = (Z×M)/(a³×N)
for fcc lattice, Z = 4
Atomic mass, M of Cu = 63.5
Here, a = 3.61 × 10-⁸ cm
ρ = (4×63.5) ÷ (3.61×10-⁸ × 6.022×10²³)
ρ = 8.96 g/cm³
(b) How many lattice points are there in one unit cell of body centered lattice ?
Ans. In bcc arrangement number of lattice point are = 8 (at corner) +1 (at centre)
Lattice points are = (8 × 1/8) + 1 = 1 + 1 = 2
Q13. “A balanced chemical equation never gives us a true picture of how a reaction takes place.” Justify the statement by taking a suitable example.
Ans. A balanced chemical equation never gives us true picture of how a reaction takes place since rarely a reaction gets completed in one step. The reactions taking place in one step are called elementary Rx. When a sequence of elementary Rx give us the product, the Rx are called complex Rx. These may be consecutive Rx, reverse Rx & side Rx.
e.g. Decomposition of H2O2 in the presence of I- in an alkaline medium
2H2O2 + I- → 2H2O + O2
Actually it occurs in two steps
H2O2 + I- → H2O + IO-
H2O2 + IO- → H2O + I- + O2
Q14. What is an adsorption isotherm ? Describe Freundlich adsorption isotherm.
Ans. The variation in amount of gas absorbed by the adsorbent with pressure at constant temp. is expresed by cure termed as adsorption isotherm.
Freundlich adsorption isotherm : Variation of extent of adsorption (x/m) with pressure at constant temp. is given by
x/m = KP 1/n ………….. (i)
Where x = mass of gas adsorbed
m = mass of adsorbent at pressure P
K and n are constants which depend upon nature of gas and adsorbent. The relationship is generally expressed in the form of curve where mass of gas adsorbed/gram of adsorbent is plotted against pressure.
Taking log of eqn. ……. (i)
log x/m = logK + 1/n logP
from slope, value of n can be calculated.
Q15. Give mechanism for following reaction :
Ans. Steps of mechanism–
Q16. Explain why :
(a) Alkyl halides, though polar, are immiscible with water ?
Ans. Alkyl halides are held together by dipoledipole forces. On the other hand H2O molecules are held together by H-bonds. When alkyl halides are added to H2O, the new force of attraction is weaker than the already existing between Rx-Rx of H2O–H2O molecules. Hence alkyl halide are immiscible in water.
(b) Grignard reagent should be prepared under anhydrous conditions ?
Ans. Grignard reagents are very reactive & reacts with moisture present in apparatus
RMgX + H2O → R–H + Mg(on) X
RMgX must be prepared under anhydrous conditions.
Q17.(a) [Fe(CN)6]⁴- and [Fe(H2O)6]²+ are of different colours in dilute solution. Why ?
Ans. In both the complexes iron is in +2 oxidation state having 3d⁶ electronic configuration having 4 unpaired electrons. But in [Fe(CN)6]⁴- these elections are paired up by CN- ligand. On the other hand they remain unpaired in [Fe(H2O)6]²+ as H2O is weak ligand. So due to difference in unpaired electrons, the complexes give different colours.
(b) Draw structures of optical isomers of [Co(NH3)Cl(en)2]²+.
Q18.(a) What is the significance of leaching in the extraction of Gold ?
Ans. Gold is leached with dilute solution of NaCN or KCN in the presence of O2 from which metal is obtained later by replacement.
4M(s) + 8CN-(aq) + 2H2O(aq) + O2(g) → 4[M(CN)2]-(aq) + 4OH-(aq) M=A4
2[M(CN)2]-(aq) + Zn(s) → [Zn(CN)4]²-(aq) + 2M(s)
(b) Why copper matte is put in silica lined converter ?
Ans. The copper matte containing Cu2S and FeS is put in silica lined converter. Some SiO2 is also added & hot air blast is blown to convert remaining FeS to FeO which is removed us slag with silica. Cu2S or CuO converted to Cu.
2FeS + O2 → 2FeO + 2SO2
FeO + SiO2 → FeSiO3
2Cu2S + 3O2 → 2Cu2O + 2SO2
2Cu2O + Cu2S → 6Cu + SO2
Q19.(a) Why does O3 act as powerful oxidizing agent ?
Ans. Ozone acts as powerful oxidizing agent because it is thermodynamically unstable and decomposes as
O3 → O2 + O
The nascent oxygen oxidizes a no. of compounds.
(b) Why does electron gain enthalpy of fluorine is less negative as compared to chlorine ?
Ans. The electron gain enthalpy of fluorine is less negative as compared to chlorine. It is due to small size of F atom. As a result there are strong inter electronic repulsions in relatively small 2p orbitals of F & thus incoming e- does not feel much attraction.
(c) Complete the reaction :
XeF2 + H2O →
Ans. 2Xe(g) + 4HF(aq) + O2(g)
Account for the following :
(a) Helium is used in diving apparatus.
Ans. Helium is used as diluent for oxygen in diving apparatus because of its very low solubility in blood.
(b) H3PO4 is tribasic acid.
Ans. H3PO4 contains three P–O–H bond. It has three ionisable hydrogen. Hence it is tribasic acid.
(c) Halogens are strong oxidizing agent.
Ans. Halogens have high electron affinities & hence have strong tendency to accept electron. That is why halogens are strong oxidizing agent.
Q20.(a) Explain giving reasons :
(i) The highest oxidation state is exhibited in oxoanions of a transition metal.
Ans. Oxoanion of transition metals have highest oxidation state e.g. O.S. of Mn in MnO4– is +7. This is because of high electro negativity of oxygen and its high oxidizing property.
(ii) The E°(M²+/M) value for copper is positive (0.34 V).
Ans. The E°(M²+/M) value for copper is positive. This show Cu is least reactive element of 1st T.S. This is because Cu has high enthalpy of atomization of enthalpy of ionization. It means I.E. required is not balanced by hydration enthalpy.
(b) What are different oxidation states exhibited by the lanthanoids ?
Ans. The principal O.S. of lanthanoids is = +3. In addition they exhibit oxidation states +2 & +4.
(a) Complete the following chemical equations :
(i) MnO4- + I- + H+ →
(ii) Cr2O7²- + Sn²+ + H+ →
(b) Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why ?
Ans. Cu exhibit +1 O.S most frequently because it has stable electronic configuration of 3d¹⁰ in +1 oxidation state.
(c) Which out of La(OH)3 and Lu(OH)3 is more basic and why ?
Ans. La(OH)3 is more basic than Lu(OH)3 due to lanthanoid contraction the size of lanthanoid ion decreases with increase in atomic no. As a result covalent character between lanthanoid ion of OH- increases.
Basic character decreases from La(OH)3 to Lu(OH)3.
Q21.(a) Give a simple chemical test to distinguish between benzaldehyde and acetophenone.
Ans. Benzaldehyde forms silver mirror with Tollen’s reagent. Acetophenone does not react.
(b) Complete the reaction :
(c) Out of benzoic acid and phenylacetic acid, which is stronger acid and why ?
Ans. Benzoic acid is stronger acid than phenyl acetic acid. This is because in benzoic acid –COOH gp is attached to SP²C of benzene ring whereas in phenylacetic acid –COOH gp is attached to SP³C.
(a) An organic compound ‘A’ (molecular formula C8H16O2) was hydrolysed with dil. H2SO4 to give a carboxylic acid ‘B’ and an alcohol ‘C’. Oxidation of ‘C’ with chromic acid produced ‘B’. ‘C’ on dehydration gives but-1-ene. Write equations for the reactions involved.
Ans. (i) Since the given compound on hydrolysis with dil. H2SO4 give carboxylic acid (B) and alcohol (C). It must be an ester.
(ii) Since the oxidation of alcohol(C) gives acid (B). Both alcohol and acid must contain same no. of carbon atoms.
(iii) Since ester contain 8 carbon atoms, acid and alcohol must contain 4 carbon each.
(iv) Alcohol (C) on dehydration gives but-1-ene, C must be straight chain alcohol i.e. butan-1-ol.
(v) (B) is obtained by oxidation of C, B must be butanoic acid.
(b) Write a short note on Wolf-Kishner reduction.
Ans. Aldehydes or Ketones are heated with hydrazine and NaOH or KOH in ethylene glycol to give hydrocarbons.
>C=O + NH2.NH2 -H2O → >C=NNH2 + KOH + glycol Heat → >CH2 + N2
Q1.(i) What is Tear gas chemically ?
Ans. (A) CCl3NO2
(ii) What is relation between [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]Cl ?
(A) Linkage isomers
(B) Coordination isomers
(C) ionisation isomers
(D) Solvate isomers
Ans. (C) ionisation isomers
(iii) In the preparation of alkyl halide from alcohol which of the following reagent is preferred ?
(A) HX + ZnCl2
Ans. (D) SO2Cl2
(iv) CH3–CH2–OH + H2SO4 (443K)→X ; what is X ?
Ans. (A) CH2=CH2
(v) Which of the following amino acids is not optically active ?
Ans. (B) Glycine
(vi) Complete the following reaction :
(vii) What are Elementary reactions ?
Ans. The reactions which takes place in one step are called elementary reactions.
(viii) What type of semiconductor is created when Ge is doped with In ?
(ix) Write the expression for relative lowering in vapour pressure.
Ans. (P⁰-P)/P⁰ = X2
(x) What is sacrificial protection ?
Ans. A method by which iron is protected from rusting by covering it with a layer of a metal more active than iron.
(xi) Expand CMC.
Ans. Critical Micelle concentration
(xii) What are the main constituents of Dettol ?
Ans. The main constituents of dettol are chloroxylenol and terpineol in suitable solvent.
(xiii) What are monomer units of Glyptal ?
Ans. Ethylene glycol and phthalic acid
(xiv) What type of linkage is there in Amylopectin ?
Ans. C1–C4 and C1–C6 glycosidic linkage
Q2. Write a short note on coupling reaction.
Ans. Benzene diazonium chloride reacts with aromatic compound having active hydrogen to form azo compounds (dyes) is called coupling reaction.
Q3. Aromatic primary amines cannot be prepared by Gabriel phthalamide reaction. Explain.
Ans. Aromatic primary amines cannot be prepared by Gabriel Phthalimide reaction because aryl halides do not undergo nucleophilic substitution with the anion formed by phthalimide.
Q4. Explain why inspite of nearly the same electronegativity, oxygen forms hydrogen bonding while chlorine does not ?
Ans. Conditions to form H-bond are small size and high electronegativity. Though chlorine has same electronegativity to that of oxygen but its size is large. That is why Cl does not form H-bonding.
Q5. Rusting of iron is said to be an electrochemical phenomenon. Give reason for this.
Ans. Rusting of iron is an electrochemical phenomenon because at a particular spot of an object made of iron, oxidation takes place and spot behave as anode & we can write reaction :
Anode 2Fe(s) → 2Fe²+ + 4e- E° Fe²+/Fe = -0.44V
Electrons released at anodic spot move through metal and go to another spot on the metal and reduce oxygen in the presence of H+. This spot behave as cathode.
Cathode O2(g) + 4H+(aq) + 4e- → 2H2O(l) E° H+/O2/H2O = 1.23V
Q6. The cell in which the following reaction occurs :
Fe³+(aq) + I-(aq) → 2Fe²+(aq) + I2(s) has 0.236 V at 298 K. Calculate the standard Gibbs energy and the equilibrium constant of the cell reaction.
Ans. E°cell = 0.236V
∆G° = -nFE°cell
∆G° = -2 × 96500C × 0.236V = -45.55 KJ
∆G° = -2.303RT logKc
log Kc = -∆G°/2.303RT = -45.55 ÷ (2.303×8.314×10-³×298) = 7.953
Kc = antilog(7.953) = 9.62×10⁷
Q7. An antifreeze solution is prepared from 222.6 g of ethylene glycol C2H4(OH)2 and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g/ml, then what shall be the molarity of the solution ?
Ans. Mass of ethylene glycol = 222.6 g
Molar mass of ethylene glycol = 62
Moles = 222.6/62 = 3.59
Molality = 3.59/200 × 1000 = 17.95 m
Moles of ethylene glycol = 3.59
Mass of solution = 200 + 222.6 = 422.6 g
Volume of solution = Mass/Density= 422.6/1.072 = 394.22 ml
Molarity = 3.59/394.22 × 1000 = 9.11 M
Q8. On the basis of osmosis explain the condition edema.
Ans. People taking a lot of salt or salty food experience water retention in tissue cells and intracellular space because of osmosis. The resulting puffiness or swelling is called edema.
Q9. Differentiate between Biodegradable and Non-biodegradable detergents. Give one example of each.
Ans. Detergents having straight hydrocarbon chains are easily decomposed by microorganisms are called biodegradable detergents e.g. sod. lauryl sulphate.
Detergents having branched hydrocarbon chain are not decomposed by microorganism and are called non biodegradable detergents e.g. sod. 4-(1,3,5,7-tetra-methyloctyl) benzene sulphonate.
Q10. What are essential and non-essential amino acids ? Give one example of each type.
Ans. The amino acids which can not be synthesized in our body and must be present in our diet are called essential amino acids e.g. valine.
On the other hand the amino acids which can be synthesized in our body are called non essential amino acids e.g. alamine.
Q11. How does the presence of double bonds in rubber molecule influence their structure and reactivity ?
Ans. Natural rubber is linear cis-1, 4 polyisoprene. The cis configuration does not allow chains to come closer due to weak intermolecular attraction. Therefore natural rubber is coiled, elastic & non crystalline.
On the other hand trans configuration occurs in Gulta-Purcha. The zig-zag chain pack more closely, hence it is crystalline, non elastic, hard of brittle.
Q12. Write the chemical reactions taking place in extraction of zinc from zinc blende.
Ans. (i) The conc. zinc blende is roasted in the presence of excess of air at 1200 K.
2ZnS + 3O2 + ∆ → 2ZnO + 2SO2
(ii) ZnO is reduced to Zn with crushed coke at 1673 K
ZnO + C → Zn + CO
(iii) Impure Zn is refined by electro refining method. The reactions taking place at cathode and anode are :
At anode Zn → Zn²+ + 2e-
At cathode Zn²+ + 2e- → Zn
Q13. Explain the splitting of ‘d’ orbitals in octahedral crystal field.
Ans. In an octahedral coordination entity the six ligands surrounding the metal atom/ion, these will be repulsion between metal d orbitals and ligands. The dx² – y² and dz² orbitals are repelled more as they are directed forwards the ligands and are raised in energy. dxy, dyz and dxz orbitals which are directed between the axes are lowered in energy. As a result degeneracy of d orbitals is removed.
Q14. How does the precipitation of colloidal smoke takes place in cottrel precipitator ?
Ans. Smoke is a colloidal solution of solid particles such as C, As compounds, dust etc. The smoke before it comes out from chimney is led through a chamber containing plates having charge opposite to smoke particles. The particles on coming in contact with these plates loose their charge and gets precipitated. The particle thus settle down on the floor of chamber.
Q15.(a) The rate constant for a first order reaction is 60 S-¹. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value ?
Ans. Let the initial conc., [A]o = a
Final conc. [A] = (1/16)×a = a/16
Rate constant = 60 S-¹
t = 2.303/k .log[A]o/[A] = 2.303/60 loga/(a/16) = 2.303/60 log16 = 2.303/60 × 1.204 = 0.046 sec
(b) What is Activation Energy ?
Ans. The energy required to form activated complex is known as activation energy.
Q16. In terms of band theory, what is difference between a conductor and a semiconductor ? Explain.
Ans. In a conductor, the energy gap between valence band and conductance band is very small or there is overlapping between valence band and conductance band. Therefore under an applied electric field the electron can jump from V. B to C. B and substance show conductivity. But in a semiconductor there is always a small energy gap between V. B and C. B.
Q17. How will you bring about following conversions ?
(a) Toluene to Benzyl alcohol
(b) But-1-ene to But-2-ene
(c) Tert-Butyl bromide to Isobutyl bromide
Q18. Write short note on :
(a) Williamson Ether Synthesis
Ans. It is an important lab. method for the prep. of symmetrical and unsymmetrical ethers. In this method an alkyl halide is allowed to react with sod. alkoxide.
R–X + R–O- –Na+ → R–O–R + NaX
(b) Kolbe’s Reaction
Ans. Sod. phenoxide react with CO2 to form O-hydroxybenzoic acid as main Rx product.
Q19.(a) Silver atom has completely filled ‘d’ orbitals (4d¹⁰) in its ground state. How can you say that it is a transition element ?
Ans. According to definition, transition elements are those which have partially filled d-orbitals in their elementary state or in their one of oxidation state. Silver can exhibit +2 O-S. in which it has partially filled d-orbital (4d^a). Hence silver is regarded as transition element.
(b) Actinoids exhibit a large number of oxidation states in comparison to corresponding members of Lanthanoid series. Why ?
Ans. Lanthanoids exhibits oxidation states of +2, +3 & +4. The reason behind this limited no. of O-S is large energy gap between 5d & 4f subshells. On the other hand actinoids show a no. of oxidation states of +3, +4, +5, +6 & +7. This is because of small energy difference between 5f, 6d & 7s orbitals.
(a) Members of second and third transition series exhibit similar atomic radii. Explain.
Ans. In third transition series 4f orbitals are filled before 5d orbital. Due to poor shielding of 4f orbitals there is regular decrease in atomic radii called lanthanoid contraction. That is why 2nd & 3rd d series members have similar atomic radii.
(b) Account for the following :
(i) Transition metals and their many compounds act as good catalysts.
Ans. Transition metals and their compounds are known for their catalytic activity. This is ascribed to their ability to adopt multiple O. States & to form complexes.
(ii) Actinoids has greater tendency to form complexes in comparison to Lanthanoids.
Ans. This is because of their high charge and small size of their ions.
Q20.(a) How will you distinguish between Benzophenone and Acetophenone with the help of a chemical test ?
Ans. Acetophenone gives yellow ppt. with alkaline solution of Iodine. Benzophenone does not respond to this test.
C6H5COCH3 + I2,NaoH → CHI3↓ + C6H5COONa
(b) Give chemical equation for the reaction of CH3COOH with following :
Ans. 3CH3COOH + PCl3 → 3CH3COCl + H3PO3
Ans. CH3COOH + LiAlH4/H3O+ → CH3CH2OH
(iii) C2H5OH in the presence of H+
Ans. CH3COOH + C2H5OH ⇌ CH3COOC2H5 + H2O
(a) Cyclohexanone forms cynohydrin in good yield but 2,2,6-trimethylcyclohexanone does not. Explain.
Ans. In 2, 2, 6 trimethyl cyclohexanone, there are three methyl gps at α-position to keto gp. Thus Nu- attack of CN- is not possible due to steric hindrance. But in case of cyclohexanone these is no such case.
(b) Explain the mechanism of nucleophilic addition reactions of aldehydes and ketones.
(c) Complete the reaction :
Q21.(a) How is SO2 an air pollutant ?
Ans. SO2 dissolves in rain water and produces acid rain. The acid rain contains H2SO4.
SO2 + ½O2 + H2O → H2SO4
(b) What is Aquaresia ? Give its reaction with Gold.
Ans. Aquaresia is three parts conc. HCl and one part conc. HNO3
Au + 4H+ + NO3– + 4Cl- → AuCl4– + NO + 2H2O
(c) Give structure of (HPO3)3
(a) Can PCl5 act as an oxidizing as well as reducing agent ? Justify.
Ans. P can show max. O.S of +5 in its compounds. In PCl5 its O.S is +5 so it can not increase its O.S beyond +5, it can not act as reducing agent. But it can act as oxidizing agent by decreasing its O. State from +5 to +3 e.g.
2Ag + PCl5 → 2AgCl + PCl3
(b) ClF3 exists but FCl3 does not. Why.
Ans. Size of chlorine (99pm) is larger than fluorine (64pm) so it can accommodate three fluorine around it. But F being smaller in size cannot accommodate three large sized chlorine.
(c) Give structure of XeF4
Q1.(i) What type of solids are electrical conductors, malleable and ductile ?
(A) Molecular solids
(B) ionic solids
(C) Metallic solids
(D) Covalent solids
Ans. (C) Metallic solids
(ii) Which of the following has highest value of Van’t Hoff factor ?
(A) 0.1M Al2(SO4)3
(B) 0.1M C6H12O6
(C) 0.1M K2SO4
(D) 0.1M NaCl
Ans. (A) 0.1M Al2(SO4)3
(iii) An example of Zero Order Reaction is :
(A) Thermal decomposition of HI on gold surface
(B) Hydrogenation of ethene
(C) Decomposition of N2O5
(D) inversion of Sucrose
Ans. (A) Thermal decomposition of HI on gold surface
(iv) What is X ?
Ans. Option (B)
(v) What type of organic compound are prepared by Gatterman-Koch reaction ?
(A) Aliphatic Aldehyde
(B) Aromatic Ketone
(C) Aliphatic Ketone
(D) Aromatic Aldehyde
Ans. (D) Aromatic Aldehyde
(vi) What is collodion ?
Ans. Collodion is a flammable syrupy solution of nitrocellulose, ether and alcohol.
(vii) Mercury cell gives a constant voltage of 1.35 V during its life time. Why ?
Ans. because the overall cell reaction does not include any ion in the solution whose concentration changes during its lifetime.
(viii) What happens when PCl5 is heated ?
Ans. PCl5 decomposes to give PCl3 and Chlorine gas, when heated. Thus, this gives an equilibrium reaction.
(ix) Give IUPAC name of [Co(NH3)4Cl(NO2)]Cl.
Ans. Tetraamminechloridonitrito-N-cobalt (III) Chloride
(x) Complete the reaction :
(xi) What are Non-reducing sugars ?
Ans. A nonreducing sugar is a carbohydrate that is not oxidized by a weak oxidizing agent (an oxidizing agent that oxidizes aldehydes but not alcohols, such as the Tollen’s reagent) in basic aqueous solution.
(xii) What are two main functions of Nucleic Acids ?
Ans. Main functions of Nucleic Acids-
(a) DNA is responsible for the transmission of inherent characters from one generation to the next. This process of transmission is called heredity.
(b) Nucleic acids (both DNA and RNA) are responsible for protein synthesis in a cell.
(xiii) Write monomer unit of Polymer .
(xiv) Why do we require artificial sweetening agents ?
Ans. We require artificial sweetening agents as they are sweet in taste but they have no food value. They go directly through the digestive tract without being digested. It helps diabetic patients to eat sweet food without adding any extra sugar in their body.
Other Questions (Coming Soon…)